Question

Prove the following by using the principle of mathematical induction for all $n\in N:(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})......(1+\frac{1}{n})=(n+1)$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})......(1+\frac{1}{n})=(n+1)$
For $n=1$, we have
$P\left(1\right):(1+\frac{1}{1})=2=(1+1)$, which is true.
Let $P\left(k\right)$ is true for some $k\in N$
$(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})......(1+\frac{1}{k})=(k+1).........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$[(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})......(1+\frac{1}{k})](1+\frac{1}{k+1})$
$=(k+1)(1+\frac{1}{k+1})$ [Using $\left(i\right)$]
$=(k+1)\left(\frac{(k+1)+1}{k+1}\right)$
$=(k+1)+1$
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$